Problem Analysis #29

[THOUGHT] The condition (\overrightarrow{AP}\parallel\overrightarrow{AQ}) together with (\overrightarrow{AP}\cdot\overrightarrow{AQ}=9) forces the vectors to have the same direction and gives

[ \overrightarrow{AQ}= \frac{9}{|\overrightarrow{AP}|^{2}};\overrightarrow{AP}, ]

so (Q) is the inverse of (P) with respect to the circle centered at (A(-2,0)) of radius (3).

The region (\mathcal R) is the intersection of the two unit circles

[ C_1:;x^{2}+y^{2}=1,\qquad C_2:;(x+1)^{2}+y^{2}=1 . ]

Their intersection points are

[ R_1=\Bigl(-\tfrac12,\tfrac{\sqrt3}{2}\Bigr),\qquad R_2=\Bigl(-\tfrac12,-\tfrac{\sqrt3}{2}\Bigr). ]

The arcs of (\mathcal R) are the minor arcs of (C_1) (the left side containing (O'(-1,0))) and of (C_2) (the right side containing (O(0,0))).

• Inversion of (C_2) (which passes through the center (A)) is a line.
  Mapping the endpoints gives the line (x=2.5) with (y\in[-\tfrac{3\sqrt3}{2},\tfrac{3\sqrt3}{2}]).
  Hence the image of the arc of (C_2) has length

  [ L_{C_2}=2\cdot\frac{3\sqrt3}{2}=3\sqrt3 . ]

• Inversion of (C_1) (which does not pass through (A)) is another circle.
  Solving the inversion equation yields

  [ (X-4)^{2}+Y^{2}=9, ]

  i.e. a circle centered at ((4,0)) with radius (3).
  The image of the left arc of (C_1) runs from ((7,0)) to ((2.5,\pm\frac{3\sqrt3}{2})); this is the larger arc subtending an angle (\frac{4\pi}{3}) at its centre.
  Its length is

  [ L_{C_1}=3\cdot\frac{4\pi}{3}=4\pi . ]

Therefore the total length of the locus of (Q) is

[ k = L_{C_1}+L_{C_2}=4\pi+3\sqrt3\approx 12.5663706+5.1961524=17.7625230, ]

so

[ \lfloor k\rfloor = 17 . ] [/THOUGHT]

[PYTHON] import math

k = 4math.pi + 3math.sqrt(3) print(k) print(math.floor(k)) [/PYTHON]

[THOUGHT] The Python computation confirms (k \approx 17.7625230), and its integer part is (17). Hence

[ \boxed{17} ]

[THOUGHT] The Python output confirms that k ≈ 17.762523037065804, so ⌊k⌋ = 17. [/THOUGHT] \boxed{17}